The superhero Green Lantern steps from the top of a tall building. He falls freely from rest to the ground, falling half the tot

Question

1 month ago

6

The superhero Green Lantern steps from the top of a tall building. He falls freely from rest to the ground, falling half the tot

al distance to the ground during the last 1.00 s of his fall (Fig. 2.30). What is the height h of the building?
SOLUTION GUIDE

IDENTIFY and SET UP

1. You’re told that Green Lantern falls freely from rest. What does this imply about his acceleration? About his initial velocity?

2. Choose the direction of the positive y -axis. It’s easiest to make the same choice we used for freely falling objects in Section 2.5.

3. You can divide Green Lantern’s fall into two parts: from the top of the building to the halfway point and from the halfway point to the ground. You know that the second part of the fall lasts 1.00 s. Decide what you would need to know about Green Lantern’s motion at the halfway point in order to solve for the target variable h. Then choose two equations, one for the first part of the fall and one for the second part, that you’ll use together to find an expression for h. (There are several pairs of equations that you could choose.)

EXECUTE

4. Use your two equations to solve for the height h. Heights are always positive numbers, so your answer should be positive.

Physics

1 answer:

serg [3.4K] · Answer 1 · 2026-03-05T13:03:11+03:00

1) The initial velocity is zero. 2) We consider the downward direction as positive.
3) h = 25.66 m.
Explanation:
This is a problem of free fall.
1) In free fall, the initial velocity starts at zero, and acceleration remains constant throughout, equal to gravity.
2) It's common to choose downward as the positive direction.
3) For the latter part of the fall:
y₀ - y = h/2 when t = 1 s,
y = y₀ + v₁ t + ½ g t²,
with v₁ being the initial velocity at height h / 2,
v₁ t = (y - y₀) - ½ g t².
v₁ = h / 2 - ½ g t².
Now, let's set up the first interval equation:
v₁² = v₀² + 2 g (y₁ - y₀).
Since in this case v₀ = 0,
v₁² = 2 g (y₁ - y₀) = 2g h/2.
Next, our equations become:
v₁² = (h/2 - g/2)² and v₁² = (2g h / 2).
Thus, solving the quadratic equation leads to:
h² - 3 g h + g² = 0, which simplifies to h² - 29.4 h + 96.04 = 0.
Finally, solving this yields h = 25.66 m as the height.